1. The problem is to find the values of $y$ for given $x$ values in the exponential functions and fill the tables with the points $(x,y)$.\n\n2. The general form of an exponential function is $y = ab^x$, where $a$ is the initial value and $b$ is the base. For each function, we substitute the $x$ values into the formula to find $y$.\n\n3. Calculate for $y = 3^x$:\n- For $x = -2$, $y = 3^{-2} = \frac{1}{3^2} = \frac{1}{9}$\n- For $x = -1$, $y = 3^{-1} = \frac{1}{3}$\n- For $x = 0$, $y = 3^0 = 1$\n- For $x = 1$, $y = 3^1 = 3$\n- For $x = 2$, $y = 3^2 = 9$\n\n4. Calculate for $y = \left(\frac{1}{3}\right)^x$:\n- For $x = -2$, $y = \left(\frac{1}{3}\right)^{-2} = 3^2 = 9$\n- For $x = -1$, $y = \left(\frac{1}{3}\right)^{-1} = 3$\n- For $x = 0$, $y = 1$\n- For $x = 1$, $y = \frac{1}{3}$\n- For $x = 2$, $y = \frac{1}{9}$\n\n5. Calculate for $y = \left(\frac{1}{4}\right)^x$:\n- For $x = -2$, $y = 4^2 = 16$\n- For $x = -1$, $y = 4$\n- For $x = 0$, $y = 1$\n- For $x = 1$, $y = \frac{1}{4}$\n- For $x = 2$, $y = \frac{1}{16}$\n\n6. Calculate for $y = 2(2)^x$:\n- For $x = -2$, $y = 2 \times 2^{-2} = 2 \times \frac{1}{4} = \frac{1}{2}$\n- For $x = -1$, $y = 2 \times \frac{1}{2} = 1$\n- For $x = 0$, $y = 2 \times 1 = 2$\n- For $x = 1$, $y = 2 \times 2 = 4$\n- For $x = 2$, $y = 2 \times 4 = 8$\n\n7. Calculate for $y = 4\left(\frac{1}{2}\right)^x$:\n- For $x = -2$, $y = 4 \times 2^2 = 4 \times 4 = 16$\n- For $x = -1$, $y = 4 \times 2 = 8$\n- For $x = 0$, $y = 4 \times 1 = 4$\n- For $x = 1$, $y = 4 \times \frac{1}{2} = 2$\n- For $x = 2$, $y = 4 \times \frac{1}{4} = 1$\n\n8. Calculate for $y = 6\left(\frac{1}{3}\right)^x$:\n- For $x = -2$, $y = 6 \times 3^2 = 6 \times 9 = 54$\n- For $x = -1$, $y = 6 \times 3 = 18$\n- For $x = 0$, $y = 6 \times 1 = 6$\n- For $x = 1$, $y = 6 \times \frac{1}{3} = 2$\n- For $x = 2$, $y = 6 \times \frac{1}{9} = \frac{2}{3}$\n\nFinal answers for the first function $y=3^x$ table points:\n$(-2, \frac{1}{9}), (-1, \frac{1}{3}), (0,1), (1,3), (2,9)$